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2 years ago

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Rotational dynamics about a fixed axis: A person pushes on a small doorknob with a force of 5.00 N perpendicular to the surface

of the door. The doorknob is located 0.800 m from axis of the frictionless hinges of the door. The door begins to rotate with an angular acceleration of 2.00 rad/s2. What is the moment of inertia of the door about the hinges

1 answer:

FrozenT [24]2 years ago

3 0

Answer:

I = 2 kgm^2

Explanation:

In order to calculate the moment of inertia of the door, about the hinges, you use the following formula:

$\tau=I\alpha$ (1)

I: moment of inertia of the door

α: angular acceleration of the door = 2.00 rad/s^2

τ: torque exerted on the door

You can calculate the torque by using the information about the Force exerted on the door, and the distance to the hinges. You use the following formula:

$\tau=Fd$ (2)

F: force = 5.00 N

d: distance to the hinges = 0.800 m

You replace the equation (2) into the equation (1), and you solve for α:

$Fd=I\alpha\\\\I=\frac{Fd}{\alpha}$

Finally, you replace the values of all parameters in the previous equation for I:

$I=\frac{(5.00N)(0.800m)}{2.00rad/s^2}=2kgm^2$

The moment of inertia of the door around the hinges is 2 kgm^2

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A plane has an average air speed (this is the speed the plane moves through air) of 750 mph. The plane flies a route of 5000 mil

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Answer:

6 hours 15 minutes

Explanation:

On the trip from L.A. to London, the plane travels at 750 mph against a headwind of 50 mph, and that makes the net 700 mph (in aviation speak, 750 is the airspeed, while 700 is the groundspeed). 5000 miles divided by 700 mph results in about 7.14 hours, or about 7 hours and 9 minutes. On the return trip, ASSUMING THE SAME WIND, the plane travels at 750 mph, but this time the wind of 50 mph is a tail wind. So the net (groundspeed) is 800 mph. Traveling 5000 miles at 800 mph only takes 6.25 hours, or 6 hours and 15 minutes.

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1 year ago

Two electrodes, separated by a distance d, in a vacuum are maintained at a constant potential difference. An electron, accelerat

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Answer:

Explanation:

Given that, the distance between the electrode is d.

The electron kinetic energy is Ek when the electrode are at distance "d" apart.

So, we want to find the K.E when that are at d/3 distance apart.

K.E = ½mv²

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Also,

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Then,

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So, as d increase K.E increase and as d decreases K.E decreases.

So,

K.E_1 / d_1 = K.E_2 / d_2

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1 year ago

Sayid made a chart listing data of two colliding objects. A 5-column table titled Collision: Two Objects Stick Together with 2 r

Alborosie

Answer:

6 m/s is the missing final velocity

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From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).

Object X had a mass of 300 kg, while object Y had a mass of 100 kg.

Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.

We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.

In numbers, and calling $P_{xi}$ the initial momentum of object X and $P_{yi}$ the initial momentum of object Y, we can derive the total initial momentum of the system: $P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}$

Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):

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We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):

$2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}$

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2 years ago

Read 2 more answers

A toy rocket engine is securely fastened to a large puck that can glide with negligible friction over a horizontal surface, take

Ad libitum [116K]

Answer:

$F=(3i+3.6j)\ N$

Explanation:

It is given that,

Mass of the puck, m = 4.8 kg

Initial velocity of the puck, $u=(1i+0j)\ m/s$

After 8 seconds, final velocity of the puck, $v=(6i+6j)\ m/s$

Let the x and y component of force is given by $F_x\ and\ F_y$ .

x component of force is given by :

$F_x=m\times \dfrac{v-u}{t}$

$F_x=4.8\times \dfrac{6-1}{8}$

$F_x=3\ N$

y component of force is given by :

$F_y=m\times \dfrac{v-u}{t}$

$F_y=4.8\times \dfrac{6-0}{8}$

$F_y=3.6\ N$

So, the component of the force is $F=(3i+3.6j)\ N$ . Hence, this is the required solution.

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2 years ago

What is the change in internal energy (in J) of a system that does 4.50 ✕ 105 J of work while 3.20 ✕ 106 J of heat transfer occu

Dmitrij [34]

Answer:

$-3.25\times 10^6 J$

Explanation:

We are given that

Work done by the system= $4.5\times 10^5$ J

Heat transfer into the system= $U_1=3.2\times 10^6$ J

Heat transfer to the environment= $U_2=6\times 10^6$ J

We have to find the change in internal energy

By first law of thermodynamics

$\Delta Q=\Delta U+w$

$\Delta Q=U_1-U_2=3.2\times 10^6-6\times 10^6=-2.8\times 10^6J$

Substitute the values then we get

$-2.8\times 10^6=\Delta U+4.5\times 10^5$

$\Delta U=-2.8\times 10^6-4.5\times 10^5=-28\times 10^5-4.5\times 10^5=-32.5\times 10^5=-3.25\times 10^6 J$

Hence, the change in internal energy = $-3.25\times 10^6 J$

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1 year ago